Chemistry - Thermodynamics - I Question with Solution | TestHub

Correct relation between change in enthalpy and change in internal energy is

$∆\mathrm{H}=∆\mathrm{U}+∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT}$

As we know,

$∆\mathrm{H}=∆\mathrm{U}+\mathrm{P}∆\mathrm{V} ....\left(\mathrm{i}\right)$

For an ideal gas

$\mathrm{PV}=\mathrm{nRT}$

For reactant, ${\mathrm{PV}}_{\mathrm{A}}={\mathrm{n}}_{\mathrm{A}}\mathrm{RT}$

For product, ${\mathrm{PV}}_{\mathrm{B}}={\mathrm{n}}_{\mathrm{B}}\mathrm{RT}$

Thus, 

$$\begin{gathered} \mathrm{P}({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}})=({\mathrm{n}}_{\mathrm{B}}-{\mathrm{n}}_{\mathrm{A}})\mathrm{RT} \\ \mathrm{P}∆\mathrm{V}=∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT} ......\left(\mathrm{ii}\right) \end{gathered}$$

Substituting (ii) in (i), we get

$\therefore ∆\mathrm{H}=∆\mathrm{U}+∆{\mathrm{n}}_{\mathrm{g}}\mathrm{RT}$